Problem: Takahashi loves sorting.

He has a permutation (p_1,p_2,...,p_N) of the integers from 1 through N. Now, he will repeat the following operation until the permutation becomes (1,2,...,N):

* First, we will define high and low elements in the permutation, as follows. The i-th element in the permutation is high if the maximum element between the 1-st and i-th elements, inclusive, is the i-th element itself, and otherwise the i-th element is low.
* Then, let a_1,a_2,...,a_k be the values of the high elements, and b_1,b_2,...,b_{N-k} be the values of the low elements in the current permutation, in the order they appear in it.
* Lastly, rearrange the permutation into (b_1,b_2,...,b_{N-k},a_1,a_2,...,a_k).



How many operations are necessary until the permutation is sorted?

Constraints

* 1 ≤ N ≤ 2×10^5
* (p_1,p_2,...,p_N) is a permutation of the integers from 1 through N.

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the number of operations that are necessary until the permutation is sorted.

Examples

Input

5
3 5 1 2 4


Output

3


Input

5
5 4 3 2 1


Output

4


Input

10
2 10 5 7 3 6 4 9 8 1


Output

6

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
 
typedef long long LL;
typedef pair<int, int> pii;
 
#define fst first
#define snd second
#define pb push_back
#define REP(i, a, b)  for(int i = (a), i##end = (b); i < i##end; ++i)
#define DREP(i, a, b) for(int i=(a-1), i##end = (b); i >=i##end; --i)
 
template <typename T> bool chkmax(T& a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> bool chkmin(T& a, T b) { return a > b ? a = b, 1 : 0; }
 
const int N = 200000;
const int oo = 0x3f3f3f3f;
 
template<typename T> T read() {
    T n(0), f(1);
    char ch = getchar();
    for(;!isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
    for(; isdigit(ch); ch = getchar()) n = n * 10 + ch - 48; 
    return n * f;
}

int n;
int a[N + 5], p[N + 5];

pii solve(int cur) {
    if(cur == n-1) {
        return p[cur] < p[cur + 1] ? pii(0, cur) : pii(1, cur + 1);
    }

    pii tmp = solve(cur + 1);

    if(!tmp.fst) {
        return p[cur] < p[tmp.snd] ? pii(0, cur) : pii(1, tmp.snd);
    }else {
        int f = tmp.snd;
        if((p[f] < p[cur] && p[cur] < p[cur + 1])
        || (p[cur] < p[cur+1] && p[cur+1] < p[f])
        || (p[cur + 1] < p[f] && p[f] < p[cur])) return tmp;

        return pii(tmp.fst + 1, cur + 1);
    }
}

int main() {
#ifdef Wearry
    freopen("data.txt", "r", stdin);
    freopen("ans.txt", "w", stdout);
#endif 

    scanf("%d", &n);
    for(int i = 1; i <= n; ++i) scanf("%d", a + i), p[a[i]] = i;
    
    if(n == 1) return puts("0"), 0;
    printf("%d\n", solve(1).fst);

    return 0;
}